Compound Inequalities

Learning Objectives

Introduction

Many times, solutions lie between two quantities, rather than continuing endlessly in one direction. For example, systolic (top number) blood pressure that is between `120` and `139` mm Hg is called borderline high blood pressure. This can be described using a compound inequality, `b<139`   and  `b>120`. Other compound inequalities are joined by the word “or.”

When two inequalities are joined by the word and, the solution of the compound inequality occurs when both inequalities are true at the same time. It is the overlap, or intersection, of the solutions for each inequality. When the two inequalities are joined by the word or, the solution of the compound inequality occurs when either of the inequalities is true. The solution is the combination, or union, of the two individual solutions.

Solving and Graphing Compound Inequalities in the Form of “or”

Let’s take a closer look at a compound inequality A statement including two inequality statements joined either by the word “or” or “and.” For example, `2x - 3` < `5` and `x + 14` > `11`. that uses or to combine two inequalities. For example, `x>6` or `x<2`. The solution to this compound inequality is all the values of `x` in which `x` is either greater than `6` or `x` is less than `2`. You can show this graphically by putting the graphs of each inequality A mathematical statement that shows the relationship between two expressions where one expression can be greater than or less than the other expression. An inequality is written by using an inequality sign (`>, <, <=, >=, !=`). together on the same number line.

A number line goes from negative 10 to positive 10. At positive 2, there is an open red circle. The number line is shaded red to the left of positive 2 and on the left end of the shaded line there is a left-facing arrow. At positive 6, there is an open blue circle. The number line is shaded blue to the right of positive 6 and on the right end of the shaded line there is a right-facing arrow.

The graph has an open circle on `6` and a blue arrow to the right and another open circle at `2` and a red arrow to the left. In fact, the only parts that are not a solution to this compound inequality are the points `2` and `6` and all the points in between these values on the number line. Everything else on the graph is a solution to this compound inequality.

Let’s look at another example of an or compound inequality, `x>3` or `x<=4`. The graph of `x>3` has an open circle on `3` and a blue arrow drawn to the right to contain all the numbers greater than `3`.

A number line goes from negative 10 to positive 10. At positive 3, there is an open blue circle. The number line is shaded blue to the right of positive 3 and on the right end of the shaded line there is a right-facing arrow.

The graph of `x<=4` has a closed circle at `4` and a red arrow to the left to contain all the numbers less than `4`.

A number line goes from negative 10 to positive 10. At positive 4, there is a filled-in red circle. The number line is shaded red to the left of positive 4 and on the left end of the shaded line there is a left-facing arrow.

What do you notice about the graph that combines these two inequalities?

A number line goes from negative 10 to positive 10. At positive 3, there is an open blue circle. The number line is shaded blue to the right of positive 3 and on the right end of the shaded line there is a right-facing arrow. At positive 4, there is a filled-in red circle. The number line is shaded red to the left of positive 4 and on the left end of the shaded line there is a left-facing arrow.

Since this compound inequality is an or statement, it includes all of the numbers in each of the solutions, which in this case is all the numbers on the number line. (The region of the line greater than `3` and less than or equal to `4` is shown in purple because it lies on both of the original graphs.) The solution to the compound inequality `x>3` or `x<=4` is the set of all real numbers!

You may need to solve one or more of the inequalities before determining the solution to the compound inequality, as in the example below.

Example

Problem

 

Solve for `x`.

`3x - 1<8` or `x - 5>0`

 

 

`x` `text()-text()5` `text()>text()` `0`
  `text()+text()5`   `text()+text()5`
`x`   `>` `5`

   

 

 

 

OR

 

`3x` `text()-text()1` `text()<text()` `8`
  `text()+text()1`   `text()+text()1`
`(3x)/3`   `<` `9/3`

 

 

 

 

 

`x<3`

 

 

`x>5` or `x<3`

 

 

Solve each inequality by isolating the variable.

 

 

 

 

 

 

 

 

 

 

 

 

Write both inequality solutions as a compound using or.

 

Answer

 

`x>5text( or )x<3` 

 
           

The solution to this compound inequality can be shown graphically.

A number line goes from negative 10 to positive 10. At positive 3, there is an open red circle. The number line is shaded red to the left of positive 3 and on the left end of the shaded line there is a left-facing arrow. At positive 5, there is an open blue circle. The number line is shaded blue to the right of positive 5 and on the right end of the shaded line there is a right-facing arrow.

Remember to apply the properties of inequality when you are solving compound inequalities. The next example involves dividing by a negative to isolate a variable.

Example

Problem

 

Solve for `y`.

`2y + 7<13` or `-3y - 2<=10`

 
`2y` `text()+text()7` `text()<text()` `13`
  `text()-text()7`   `text()-text()7`
`(2y)/2`   `text()<text()` `6/2`

 

 

 

 

 

`y<3`

 

OR

 

`text()-text()3y` `text()-text()2` `text()<=text()` `10`
  `text()+text()2`   `text()+text()2`
`(-3y)/-3`   `text()>=text()` `12/-3`

 

 

 

 

`y>=-4`

 

`y<3` or `y>=-4`

 

Solve each inequality separately.

 

 

 

 

 

 

 

 

The inequality sign is reversed with division by a negative number.

 

 

 

Since `y` could be less than `3` or greater than or equal to `-4`, `y` could be any number.

Answer

The solution is all real numbers.
         

This number line shows the solution set of `y<3` or `y>=-4`.

A number line goes from negative 10 to positive 10. At negative 4, there is a filled-in blue circle. The number line is shaded blue to the right of negative 4 and on the right end of the shaded line there is a right-facing arrow. At positive 3, there is an open red circle. The number line is shaded red to the left of positive 3 and on the left end of the shaded line there is a left-facing arrow.

Example

Problem

 

Solve for `z`.

`5z-3> -18 text( or)-2z-1>15`

 
`5z` `text()-text()3` `text()>text()` `text()-text()18`
  `text()+text()3`   `text()+text()3`
`(5z)/5`   `>` `(-15)/5`

 

 

 

 

 

`z> -3`

OR

 

`text()-text()2z` `text()-text()1` `text()>text()` `15`
  `text()+text()1`   `text()+text()1`
`(-2z)/-2`   `<` `16/-2`

 

 

 

 

 

`z<-8`

 

`z> -3 text( or )z<-8`

 

Solve each inequality separately.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Combine the solutions.

 

 

Answer

`z> -3text( or )z<-8`

 
         

This number line shows the solution set of `z> -3text( or )z<-8` .

A number line goes from negative 10 to positive 10. At negative 8, there is an open red circle. The number line is shaded red to the left of negative 8 and on the left end of the shaded line there is a left-facing arrow. At negative 3, there is an open blue circle. The number line is shaded blue to the right of negative 3 and on the right end of the shaded line there is a right-facing arrow.

Solve for `h`.

`h + 3>12text( or )3-2h>9`

 

A) `h<3text( or )h> -3` 

 

B) `h>9text( or )h> -3`

 

C) `h> -9text( or )h<3`

 

D) `h>9text( or )h<-3`

 

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Solving and Graphing Compound Inequalities in the Form of “and”

The solution of a compound inequality that consists of two inequalities joined with the word and is the intersection of the solutions of each inequality. In other words, both statements must be true at the same time. The solution to an and compound inequality are all the solutions that the two inequalities have in common. Graphically, you can think about it as where the two graphs overlap.

Think about the example of the compound inequality: `x<5` and  `x>=-1`. The graph of each individual inequality is shown in color.

A number line goes from negative 10 to positive 10. At positive 5, there is an open red circle. The number line is shaded red to the left of positive 5 and on the left end of the shaded line there is a left-facing arrow. The formula x is less than 5 appears above the shaded line. At negative 1, there is a filled-in blue circle. The number line is shaded blue to the right of negative 1 and on the right end of the shaded line there is a right-facing arrow. The formula x is greater than or equal to negative 1 appears above the shaded line.

Since the word and joins the two inequalities, the solution is the overlap of the two solutions. This is where both of these statements are true at the same time.

The solution to this compound inequality is shown below.

A number line goes from negative 10 to positive 10. At negative 1, there is a filled-in blue circle. The number line is shaded to the right of negative 1 and ends with an open red circle at positive 5. The solution to the compound inequality, negative 1 is less than or equal to x is less than or equal to 5, appears above the shaded line.

Notice that in this case, you can rewrite `x>=-1` and `x<5` as `-1<=x<5` since the solution is between `-1` and `5`, including `-1`. You read `-1<=x<5` as “`x` is greater than or equal to `-1` and less than `5`.” You can rewrite an and statement this way only if the answer is between two numbers.

Let’s look at two more examples.

Example

Problem

 

Solve for `x`.

`1-4x<=21text( and )5x+2>=22`
`1` `text()-text()4x` `text()<=text()` `21`
`text()-text()1`     `text()-text()1`
  `(-4x)/-4` `text()>=text()` `20/-4`

 

 

 

 

 

`x>=-5`

 

AND

 

 

`5x` `text()+text()2` `text()>=text()` `22`
  `text()-text()2`   `text()-text()2`
`(5x)/5`   `text()>=text()` `20/5`

 

 

 

 

 

`x>=4`

 

 

`x>=-5text( and )x>=4`

 

Solve each inequality for `x`.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Determine the intersection of the solutions.

The number line below shows the graphs of the two inequalities in the problem. The solution to the compound inequality is `x>=4`, as this is where the two graphs overlap.

A number line goes from negative 10 to positive 10. At negative 5, there is a filled-in blue circle. The number line is shaded blue to the right of negative 5 and on the right end of the shaded line there is a right-facing arrow. The formula x is greater than or equal to negative 5 appears above the shaded line. At positive 4, there is a filled-in red circle. The number line is shaded red to the right of positive 4 and on the right end of the shaded line there is a right-facing arrow. The formula x is greater than or equal to 4 appears above the shaded line.

And the solution can be represented as: A number line goes from negative 10 to positive 10. At positive 4, there is a filled-in circle. The number line is shaded to the right of positive 4 and on the right end of the shaded line there is a right-facing arrow. The solution to the compound inequality, x is greater than or equal to 4, appears above the shaded line.

 

Answer

 

`x>=4`

 
         

Example

Problem

 

Solve for `x`.

`5x-2<=3text( and )4x+7>3`

 
`5x` `text()-text()2` `text()<=text()` `3`
  `text()+text()2`   `text()+text()2`
`(5x)/5`   `<=` `5/5`

 

 

 

 

 

`x<=1`

 

AND

 

`4x` `text()+text()7` `text()>text()` `3`
  `text()-text()7`   `text()-text()7`
`(4x)/4`   `>` `(-4)/4`

 

 

 

 

 

`x> -1`

 

`x<=1text( and )x> -1`

 

Solve each inequality separately.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Find the overlap between the solutions.

 

 

The two inequalities can be represented graphically as:

 

A number line goes from negative 10 to positive 10. At positive 1, there is a filled-in red circle. The number line is shaded red to the left of positive 1 and on the left end of the shaded line there is a left-facing arrow. The formula x is less than or equal to 1 appears above the shaded line. At negative 1, there is an open blue circle. The number line is shaded blue to the right of negative 1 and on the right end of the shaded line there is a right-facing arrow. The formula x is greater than negative 1 appears above the shaded line.

And the solution can be represented as:

 

A number line goes from negative 10 to positive 10. At positive 1, there is a filled-in circle. The number line is shaded to the left of positive 1 and ends with an open circle at negative 1. The solution to the compound inequality, negative 1 is less than x is less than or equal to 1, appears above the shaded line.

Answer

 

`x<=1text( and )x> -1`

 
         

Rather than splitting a compound inequality in the form of `a<x<b` into two inequalities `x<b` and `x>a`, you can more quickly solve the inequality by applying the properties of inequality to all three segments of the compound inequality. Two examples are provided below.

Example

Problem

 

Solve for `x` . 

`3<2x+3<=7`

 
`3` `text()<text()` `2x` `text()+text()3` `text()<=text()` `7`
`text()-text()3`     `text()-text()3`   `text()-text()3`
`0/2` `<` `(2x)/2`   `<=` `4/2`

 

 

 

 

 

`0<x<=2`

 

Isolate the variable by subtracting `3` from all `3` parts of the inequality, and then dividing each part by `2`.

 

 

 

Answer

`0<x<=2`

 
           

Example

Problem

 

Solve for `x`.

`-3<=2x+7<11`

 
`text()-text()3` `text()<=text()` `2x` `text()+text()7` `text()<text()` `11`
`text()-text()7`     `text()-text()7`   `text()-text()7`
`(-10)/2` `text()<=text()` `(2x)/2`   `text()<text()` `4/2`

 

 

 

 

 

`-5<=x<2`

 

Isolate the variable by subtracting `7` from all `3` parts of the inequality, and then dividing each part by `2`.

 

 

 

 

Answer

`-5<=x<2`

 
           

To solve inequalities like `a<x<b`, use the addition and multiplication properties of inequality to solve the inequality for `x`. Whatever operation you perform on the middle portion of the inequality, you must also perform to each of the outside sections as well. Pay particular attention to division or multiplication by a negative.

Which of the following compound inequalities represents the graph on the number line below?

 

A number line goes from negative 10 to positive 10. At negative 1, there is an open circle. The number line is shaded to the left of negative 1 and ends with a filled-in circle at negative 8.

 

A) `-8>=x> -1`

 

B) `-8<=x<-1`

 

C) `-8<=x> -1`

 

D) `-8>=x<-1`

 

+ 

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Special Cases of Compound Inequalities

The solution to a compound inequality with and is always the overlap between the solution to each inequality. There are three possible outcomes for compound inequalities joined by the word and:

`1`. The solution could be all the values between two endpoints.

A number line goes from negative 10 to positive 10. At positive 1, there is a filled-in red circle. The number line is shaded red to the left of positive 1 and on the left end of the shaded line there is a left-facing arrow. The formula x is less than or equal to 1 appears above the shaded line. At negative 1, there is an open blue circle. The number line is shaded blue to the right of negative 1 and on the right end of the shaded line there is a right-facing arrow. The formula x is greater than negative 1 appears above the shaded line.

A number line goes from negative 10 to positive 10. At positive 1, there is a filled-in red circle. The number line is shaded to the left of positive 1 and ends with an open blue circle at negative 1. The formula negative 1 is less than x is less than or equal to 1 appears above the shaded line.

`2`. The solution could begin at a point on the number line and extend in one direction.

A number line goes from negative 10 to positive 10. At negative 3, there is an open blue circle. The number line is shaded blue to the right of negative 3 and on the right end of the shaded line there is a right-facing arrow. The formula x is greater than negative 3 appears above the shaded line. At positive 4, there is a filled-in red circle. The number line is shaded red to the right of positive 4 and on the right end of the shaded line there is a right-facing arrow. The formula x is greater than or equal to 4 appears above the shaded line.

A number line goes from negative 10 to positive 10. At positive 4, there is a filled-in circle. The number line is shaded to the right of positive 4 and on the right end of the shaded line there is a right-facing arrow. The solution, x is greater than or equal to 4, appears above the shaded line.

`3`. In cases where there is no overlap between the two inequalities, there is no solution to the compound inequality.

A number line goes from negative 10 to positive 10. At negative 3, there is an open red circle. The number line is shaded red to the left of negative 3 and on the left end of the shaded line there is a left-facing arrow. The formula x is less than negative 3 appears above the shaded line. At positive 3, there is an open blue circle. The number line is shaded blue to the right of positive 3 and on the right end of the shaded line there is a right-facing arrow. The formula x is greater than 3 appears above the shaded line.

An example is shown below.

Example

Problem

 

Solve for `x`.

`x+2>5` and  `x+4<5`

 
`x` `text()+text()2` `text()>text()` `5`
  `-2`   `text()-text()2`
`x`   `text()>text()` `3`

 

 

 

 

AND

 

`x` `text()+text()4` `text()<text()` `5`
  `text()-text()4`   `text()-text()4`
`x`   `text()<text()` `1`

 

 

 

 

`x>3text( and )x<1`

 

Solve each inequality separately.

 

 

 

 

 

 

 

 

Find the overlap between the solutions.

 

A number line goes from negative 10 to positive 10. At positive 1, there is an open red circle. The number line is shaded red to the left of positive 1 and on the left end of the shaded line there is a left-facing arrow. The formula x is less than 1 appears above the shaded line. At positive 3, there is an open blue circle. The number line is shaded blue to the right of positive 3 and on the right end of the shaded line there is a right-facing arrow. The formula x is greater than 3 appears above the shaded line.

Answer  There is no overlap between `x>3text( and )x<1`, so there is no solution.

 
         

Summary

A compound inequality is a statement of two inequality statements linked together either by the word or or by the word and. Sometimes, an and compound inequality is shown symbolically, like `a<x<b`, and does not even need the word and. Because compound inequalities represent either a union or intersection of the individual inequalities, graphing them on a number line can be a helpful way to see or check a solution. Compound inequalities can be manipulated and solved much the same way any inequality is solved, paying attention to the properties of inequalities and the rules for solving them.